Given $f(n)=\sqrt{2n+1+2\sqrt{{n}^{2}+n}}$, Evaluate $\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+\cdots +\frac{1}{f(99)}$

Alfredeim
2022-09-12
Answered

Given $f(n)=\sqrt{2n+1+2\sqrt{{n}^{2}+n}}$, Evaluate $\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+\cdots +\frac{1}{f(99)}$

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Holly Schmidt

Answered 2022-09-13
Author has **10** answers

The hint.

Use

$$\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}$$

and the telescopic sum.

Finally, we obtain

$$\sqrt{100}-1=9.$$

Use

$$\frac{1}{\sqrt{n+1}+\sqrt{n}}=\sqrt{n+1}-\sqrt{n}$$

and the telescopic sum.

Finally, we obtain

$$\sqrt{100}-1=9.$$

lemondedeninaug

Answered 2022-09-14
Author has **2** answers

Very simple Hint: Rationalize the denominator

$$\frac{1}{f(x)}=\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n}-\sqrt{n+1}}{(\sqrt{n}+\sqrt{n+1})(\sqrt{n}-\sqrt{n+1})}=\sqrt{n+1}-\sqrt{n}$$

$$\frac{1}{f(x)}=\frac{1}{\sqrt{n}+\sqrt{n+1}}=\frac{\sqrt{n}-\sqrt{n+1}}{(\sqrt{n}+\sqrt{n+1})(\sqrt{n}-\sqrt{n+1})}=\sqrt{n+1}-\sqrt{n}$$

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But when being asked to change this number into a fraction, I do not understand where the $\frac{6}{7}$ comes from in the answer of $4\frac{6}{7}$. Can someone please help me understand?

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But when being asked to change this number into a fraction, I do not understand where the $\frac{6}{7}$ comes from in the answer of $4\frac{6}{7}$. Can someone please help me understand?

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Find $\frac{y}{x}$ from $3x+3y=yt=xt+2.5x$ I need to find the ratio of

$\frac{y}{x}$

If given that

$3x+3y=yt=xt+2.5x$

So what I tried is:

$t=\frac{3x+3y}{y}$

And then put it in the equation

$\frac{x(3x+3y)}{y}+2.5x=\frac{(3x+3y)}{y}y$

$\frac{x(3x+3y)}{y}+2.5x=3x+3y$

$\frac{3{x}^{2}}{y}+\frac{3yx}{y}+2.5x=3x+3y$

$\frac{3{x}^{2}}{y}+3x+2.5x=3x+3y$

$\frac{3{x}^{2}}{y}+2.5x=3y$

Here I got stuck. I didn't know how to find the ratio. Can someone help me?

$\frac{y}{x}$

If given that

$3x+3y=yt=xt+2.5x$

So what I tried is:

$t=\frac{3x+3y}{y}$

And then put it in the equation

$\frac{x(3x+3y)}{y}+2.5x=\frac{(3x+3y)}{y}y$

$\frac{x(3x+3y)}{y}+2.5x=3x+3y$

$\frac{3{x}^{2}}{y}+\frac{3yx}{y}+2.5x=3x+3y$

$\frac{3{x}^{2}}{y}+3x+2.5x=3x+3y$

$\frac{3{x}^{2}}{y}+2.5x=3y$

Here I got stuck. I didn't know how to find the ratio. Can someone help me?

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I understand I have to rewrite the fraction somehow for the denominator not to equal 0, but I don't know where to start.

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